Question

Estimate the change in enthalpy and entropy when liquid ammonia at 270K is compressed from its saturation pressure of 38/ kPato 1200 Wa. For saturated liquid ammonia at 270K,Vl=1.551×10−3m3kg−1, and β=2.095×10−3K−1.

Answer 1

The change in enthalpy and entropy when liquid ammonia at 270 K is compressed from its saturation pressure of 381  KPa to 1200 KPa (in the question / and w are the typo mistakes) will be -2.66 × 10⁻³ KJ/Kg.K

Initial volume (V1) = 1.551 × 10⁻³ m³Kg⁻¹

     β = 2.095 × 10⁻³K⁻¹

     ΔH = CpΔT + [V-T(ΔV/ΔT)]ΔP

     ΔS = CpΔT - (ΔV/ΔT)p ΔP

For isothermal conditions

    ΔH =  [V-T(ΔV/ΔT)]p ΔP

    ΔH = V - (1 - βT) ΔP

Let V and β constant

    ΔH = V (1 - βT) ΔP

    ΔH = 1.551 × 10⁻³ m³Kg⁻¹ [1 - ( 2.095 × 10⁻³  × 270)] (1200 - 381)

   ΔH = 0.557 KJ/Kg

Now calculate dS

    ΔS = -(ΔV/ΔT)p ΔP = - βV ΔP

    ΔS = βV ΔP

    ΔS = -1.55 × 10⁻³ × 2.095 × 10⁻³ (1200 -381)

    ΔS = -2.66 × 10⁻³ KJ/Kg.K

You can also learn about  enthalpy and entropy from the following question:

brainly.com/question/15094832

#SPJ4